Find the equation of the plane containing a line: x + 1 2 = y + 1 - 1 = z - 3 4 and perpendicular to the plane: x + 2 y + z = 12

Equation of the given line: x + 1 2 = y + 1 - 1 = z - 3 4 ⇒ ( x 1 , y 1 , z 1 ) = ( - 1 , - 1 , 3 ) ⇒ ( a 1 , b 1 , c 1 ) = ( 2 , - 1 , 4 ) Equation of the given plane: x + 2 y + z = 12 ⇒ ( a 2 , b 2 , c 2 ) = ( 1 , 2 , 1 ) The required plane is: x - x 1 y - y 1 z - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 0 ⇒ x + 1 y + 1 z - 3 2 - 1 4 1 2 1 = 0 ⇒ ( x + 1 ) ( - 1 - 8 ) - ( y + 1 ) ( 2 - 4 ) + ( z - 3 ) ( 4 + 1 ) = 0 ⇒ - 9 ( x + 1 ) + 2 ( y + 1 ) + 5 ( z - 3 ) = 0 ⇒ - 9 x - 9 + 2 y + 2 + 5 z - 15 = 0 ⇒ - 9 x + 2 y + 5 z - 22 = 0 ⇒ 9 x - 2 y - 5 z + 22 = 0

Find the equation of the plane containing a line: x + 1 2 = y + 1 - 1 = z - 3 4 and perpendicular to the plane: x + 2 y + z = 12

Equation of the given line: x + 1 2 = y + 1 - 1 = z - 3 4 ⇒ ( x 1 , y 1 , z 1 ) = ( - 1 , - 1 , 3 ) ⇒ ( a 1 , b 1 , c 1 ) = ( 2 , - 1 , 4 ) Equation of the given plane: x + 2 y + z = 12 ⇒ ( a 2 , b 2 , c 2 ) = ( 1 , 2 , 1 ) The required plane is: x - x 1 y - y 1 z - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 0 ⇒ x + 1 y + 1 z - 3 2 - 1 4 1 2 1 = 0 ⇒ ( x + 1 ) ( - 1 - 8 ) - ( y + 1 ) ( 2 - 4 ) + ( z - 3 ) ( 4 + 1 ) = 0 ⇒ - 9 ( x + 1 ) + 2 ( y + 1 ) + 5 ( z - 3 ) = 0 ⇒ - 9 x - 9 + 2 y + 2 + 5 z - 15 = 0 ⇒ - 9 x + 2 y + 5 z - 22 = 0 ⇒ 9 x - 2 y - 5 z + 22 = 0

Show that the equation of the plane which contains the parallel lines x − 4 1 = y − 3 − 4 = z − 2 5 and x − 3 1 = y + 2 − 4 = z 5 is 11 x − y − 3 z = 35 .

Given that the equation of lines x − 4 1 = y − 3 − 4 = z − 2 5 and x − 3 1 = y + 2 − 4 = z 5 We know that the equation of the plane containing these lines x - x 1 a 1 = y − y 1 b 1 = z - z 1 c 1 and x - x 2 a 2 = y − y 2 b 2 = z - z 2 c 2 is x - x 1 y - y 1 z - z 1 x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 = 0 Here, x 2 = 3 , x 1 = 4 ; y 2 = - 2 , y 1 = 3 ; z 2 = 0 , z 1 = 2 a 1 = 1 , b 1 = - 4 , c 1 = 5 ; a 2 = 1 , b 2 = - 4 , c 2 = 5 ∴ The equation of the plane containing both these lines is x - 4 y - 3 z - 2 - 1 - 5 - 2 1 - 4 5 = 0 ⇒ x - 4 - 25 - 8 - y - 3 - 5 + 2 + z - 2 4 + 5 = 0 ⇒ - 33 x + 3 y + 9 z - 105 = 0 ⇒ 11 x − y − 3 z = 105 3 = 35 Hence, the required equation of plane is 11 x − y − 3 z = 35 .

Show that the equation of the plane through the line x l = y m = z n and perpendicular to plane containing the lines and x n = y l = z m is x m 2 - l n + y n 2 - l m + z l 2 - n m = 0 .

Given that: the lines x l = y m = z n and x n = y l = z m We know that the equation of the plane containing these lines x - x 1 a 1 = y − y 1 b 1 = z - z 1 c 1 and x - x 2 a 2 = y − y 2 b 2 = z - z 2 c 2 is x - x 1 y - y 1 z - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 0 . Here, x 2 = 0 , x 1 = 0 ; y 2 = 0 , y 1 = 0 ; z 2 = 0 , z 1 = 0 a 1 = l , b 1 = m , c 1 = n ; a 2 = n , b 2 = l , c 2 = m The equation of the plane containing both these lines is x - 0 y - 0 z - 0 l m n n l m = 0 ⇒ x - 0 m 2 - l n - y - 0 l m - n 2 + z - 0 l 2 - n m = 0 ⇒ x m 2 - l n + y n 2 - l m + z l 2 - n m = 0 Hence, the required equation of plane is x m 2 - l n + y n 2 - l m + z l 2 - n m = 0 . Hence, Proved.